Computer Science Fundamentals: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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1Formulas and methods

Number systems

Place values
binary 2k, octal 8k, hex 16k
To decimal
Σ dk·bk
From decimal
divide by the base, read the remainders bottom-up
Hex ↔ binary
one hex digit = 4 bits
Two’s complement
invert the bits and add 1

Propositional logic & proofs

Rows of a truth table
2ⁿ
Implication
p → q ≡ ¬p ∨ q ≡ ¬q → ¬p
Negated implication
¬(p → q) ≡ p ∧ ¬q
Biconditional
p ↔ q ≡ (p → q) ∧ (q → p)
De Morgan
¬(p ∧ q) ≡ ¬p ∨ ¬q, ¬(p ∨ q) ≡ ¬p ∧ ¬q
Distributive laws
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r), p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Negated quantifiers
¬∀x P(x) ≡ ∃x ¬P(x), ¬∃x P(x) ≡ ∀x ¬P(x)
Modus ponens, modus tollens
p → q, p ⊢ q; p → q, ¬q ⊢ ¬p

Method

  1. Count the different variables: the truth table has 2ⁿ rows, starting with all of them 1.
  2. Give every subformula its own column and fill the columns from the inside out.
  3. Apply the rule of each connective: → is 0 only for 1 → 0, ↔ is 1 when both sides agree.
  4. For equivalences, compare the final columns row by row; for an argument, look only at the rows in which every premise is 1.
  5. To negate a quantified statement, move ¬ inwards, swap ∀ and ∃, and negate the inner formula.

Digital logic: gates, K-maps & flip-flops

Basic gates
A·B, A + B, A′, (A·B)′, (A + B)′, A ⊕ B
Boolean rules
X + X′ = 1, X·X′ = 0, X + XY = X, XY + XY′ = X
De Morgan
(X·Y)′ = X′ + Y′, (X + Y)′ = X′·Y′
Karnaugh map order
00, 01, 11, 10 (Gray code)
Group of 2ᵏ cells
loses k variables
4-to-1 multiplexer
F = A′B′I0 + A′BI1 + AB′I2 + ABI3
Flip-flops
D: Q⁺ = D, T: Q⁺ = T ⊕ Q, JK: Q⁺ = JQ′ + K′Q
Counters
n-bit: 2ⁿ states; mod-N: smallest n with 2ⁿ ≥ N; fout = fclk/N
Ripple counter
fmax = 1/(n·tpd)
Ring and Johnson counters
n and 2n states

Method

  1. For a circuit, work from the inputs to the output one gate at a time and write the value on every wire, or write its expression and fill in the truth table.
  2. For simplification, put the 1s (and the don’t-cares as X) into a Karnaugh map with Gray-code rows and columns: 00, 01, 11, 10.
  3. Cover the 1s with as few and as large groups of 1, 2, 4, 8 or 16 cells as possible, wrapping round the edges; each group is one product term made of the variables that stay constant. Group the 0s for the POS form.
  4. For flip-flops and counters, read the inputs at each active clock edge, write the state after every edge, and remember that an n-bit counter wraps round after 2ⁿ states.

Number theory & modular arithmetic

Division with remainder
a = q·n + r, 0 ≤ r < n
Euclid
gcd(a, b) = gcd(b, a mod b)
Least common multiple
lcm(a, b) = a·b/gcd(a, b)
Inverse
a·a⁻¹ ≡ 1 (mod n), gcd(a, n) = 1
Linear congruence
a·x ≡ b → x ≡ a⁻¹·b (mod n)
Euler’s totient
φ(n) = n·∏(1 − 1/p), φ(pᵏ) = pᵏ − pᵏ⁻¹
Fermat
aᵖ⁻¹ ≡ 1 (mod p)
RSA
n = pq, φ = (p − 1)(q − 1), d = e⁻¹ mod φ, c = mᵉ mod n, m = cᵈ mod n

Method

  1. Reduce every number modulo n first; remainders lie between 0 and n − 1.
  2. For a gcd, divide repeatedly with Euclid’s algorithm; the last nonzero remainder is the gcd.
  3. For an inverse, run the extended algorithm and keep the coefficient of a; add n if it is negative.
  4. Solve a·x ≡ b by multiplying with a⁻¹; combine two congruences by writing x = r₁ + m₁k.
  5. For powers, write the exponent in binary and square repeatedly; in RSA, d = e⁻¹ mod φ(n) with φ(n) = (p − 1)(q − 1).

Graph theory

Handshake lemma
Σ deg(v) = 2|E|
Complete graphs
|E(Kn)| = n(n − 1)/2, |E(Km,n)| = m·n
Regular graph
|E| = n·k/2
Tree and forest
|E| = n − 1, |E| = n − c
Leaves of a tree
L = 2 + Σ (deg(v) − 2) over deg(v) ≥ 3
Euler
circuit: all degrees even; path: exactly 2 odd degrees
Hamilton circuits in Kn
(n − 1)!/2
Dijkstra update
d(v) = min(d(v), d(u) + w(u, v))
Walks of length L
(Mᴸ)ij

Method

  1. Write down the degree of every vertex; the degrees add up to twice the number of edges.
  2. For Euler questions, check that the graph is connected and count the vertices with odd degree.
  3. For trees, use n − 1 edges and the handshake lemma.
  4. For Dijkstra, keep a table of tentative distances and settle the smallest unsettled vertex each round.
  5. For a minimum spanning tree, sort the edges by weight and skip every edge that closes a cycle (Kruskal), or grow one tree from a start vertex (Prim).

IPv4 subnetting

Hosts per subnet
2(32 − prefix) − 2 (/31: 2, /32: 1)
Block size
256 − (mask octet)
Network address
IP AND mask
Broadcast address
IP OR (NOT mask): all host bits set to 1
Subnets from n borrowed bits
2n

2Practice problems

1.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −25 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −258 bit
U =
2.Number systemsHard
2677 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 2677 items its own binary code of the same length. What is the smallest number of bits b needed?

2677
b = bit
3.Propositional logic & proofsMedium
pqr¬r(p → q) ∨ ¬r(p ∨ (r ∨ q)) → ((p → q) ∨ ¬r)111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is (p ∨ (r ∨ q)) → ((p → q) ∨ ¬r) true? Call this number N.

(p ∨ (r ∨ q)) → ((p → q) ∨ ¬r)
N =
4.Propositional logic & proofsHard
p¬rrr ∧ ¬p

The premises (q ∧ r) → p; q; ¬p are given. Which conclusion follows from them? Click it.

(q ∧ r) → pq¬p
  1. 1p
  2. 2¬r
  3. 3r
  4. 4r ∧ ¬p
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110010011031214150716

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(1, 2, 4, 5, 6). How many product terms N does it have?

F(A, B, C) = Σm(1, 2, 4, 5, 6)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKRING

A 4-bit ring counter Q₃Q₂Q₁Q₀ starts at 1100. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 4 clock edges.

Q₃Q₂Q₁Q₀ = 11004
Q =
7.Number theory & modular arithmeticMedium

Solve 65x ≡ 88 (mod 97). Give the solution x with 0 < x < 97.

65x ≡ 88 (mod 97)
x =
8.Number theory & modular arithmeticHard
39c = me​ mod n?m = cd​ mod nm(n, e) = (187, 17)d

Encrypt the message m = 39 with the RSA public key (n, e) = (187, 17). Find the ciphertext c.

n = 187e = 17m = 39
c =
9.Algorithms & complexityMedium

What is the time complexity of the standard dynamic programming algorithm for the longest common subsequence of strings of lengths m and n?

  1. AO(mn)
  2. BO(mn log n)
  3. CO(m + n)
  4. DO(2m+n)
10.Algorithms & complexityHard

A problem is NP-complete if it is:

  1. ASolvable in polynomial time
  2. BOutside NP but harder than all NP problems, such as the halting problem
  3. CIn NP and every NP problem reduces to it in polynomial time
  4. DImpossible to solve
11.Data structuresMedium

Why is traversing an array often much faster in practice than traversing a linked list of the same length?

  1. ALinked lists always store elements on disk
  2. BArray traversal is O(1) in total, while linked list traversal is O(n)
  3. CArrays use fewer instructions to compare values
  4. DArray elements are contiguous, so they make good use of the CPU cache
12.Data structuresHard

A dynamic array doubles when full. Why do many implementations shrink it only when it is one quarter full, not half full?

  1. ABecause halving at one half would lose elements
  2. BBecause the memory allocator can only release memory in quarters of a block
  3. CTo avoid repeated resizing when pushes and pops alternate near the boundary
  4. DTo keep the array sorted
13.Graph theoryMedium

A tree has 2 vertices of degree 3, 2 vertices of degree 4, 4 vertices of degree 5, and all its other vertices are leaves (degree 1). How many leaves L does it have?

2 × deg 32 × deg 44 × deg 5
L =
14.Graph theoryHard
ABCDEF

How many closed walks of length 3 start and end at E? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =
15.Databases & SQLMedium

In relational algebra, the projection operator (π) corresponds to:

  1. AChoosing which rows satisfy a given condition
  2. BCombining two tables on a key
  3. CChoosing which columns appear in the result
  4. DSorting the result
16.Databases & SQLHard

Which SQL feature is best suited to finding every employee below a manager in an organisation tree of unknown depth?

  1. AA recursive common table expression (WITH RECURSIVE)
  2. BA GROUP BY on manager_id
  3. CA single self-join of the employees table on manager_id
  4. DA UNIQUE constraint on manager_id
17.Computer networksMedium

How many usable host addresses does a /24 IPv4 network have?

  1. A255
  2. B254
  3. C252
  4. D256
18.Computer networksHard

Head-of-line blocking in HTTP/2 over TCP occurs because:

  1. AServers cannot handle parallel requests
  2. BEach request needs its own new TCP connection, so later requests wait in a queue
  3. CA lost TCP segment delays all multiplexed streams until it is retransmitted
  4. DHTTP/2 uses UDP
19.IPv4 subnettingMedium
/24 → /2511111111111111111111111110000000network bitshost bits

The network 192.168.243.0/24 is divided into /25 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 2?

192.168.243.0/24/25subnet 2 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard

The network 192.168.73.0/24 is split with VLSM for these LANs: A: 34 hosts, B: 9 hosts, C: 6 hosts, D: 19 hosts, E: 12 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN D?

A: 34B: 9C: 6D: 19E: 12
x =

3Solutions

1.Number systems

Answer: U = 231

Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.

Invert and add one

25 = 0001 1001
inverted: 1110 0110
+1 → 1110 0111

Read it as unsigned

1110 0111₂ = 256 − 25 = 231

Answer

U = 231
2.Number systems

Answer: b = 12 bit

Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.

Powers of two

211 = 2048 < 2677
212 = 4096 ≥ 2677

Answer

b = 12
3.Propositional logic & proofs

Answer: N = 7

Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.

Truth table row by row

p = 1, q = 1, r = 1: r ∨ q = 1, p ∨ (r ∨ q) = 1, p → q = 1, ¬r = 0, (p → q) ∨ ¬r = 1, (p ∨ (r ∨ q)) → ((p → q) ∨ ¬r) = 1
p = 1, q = 1, r = 0: r ∨ q = 1, p ∨ (r ∨ q) = 1, p → q = 1, ¬r = 1, (p → q) ∨ ¬r = 1, (p ∨ (r ∨ q)) → ((p → q) ∨ ¬r) = 1
p = 1, q = 0, r = 1: r ∨ q = 1, p ∨ (r ∨ q) = 1, p → q = 0, ¬r = 0, (p → q) ∨ ¬r = 0, (p ∨ (r ∨ q)) → ((p → q) ∨ ¬r) = 0
p = 1, q = 0, r = 0: r ∨ q = 0, p ∨ (r ∨ q) = 1, p → q = 0, ¬r = 1, (p → q) ∨ ¬r = 1, (p ∨ (r ∨ q)) → ((p → q) ∨ ¬r) = 1
p = 0, q = 1, r = 1: r ∨ q = 1, p ∨ (r ∨ q) = 1, p → q = 1, ¬r = 0, (p → q) ∨ ¬r = 1, (p ∨ (r ∨ q)) → ((p → q) ∨ ¬r) = 1
p = 0, q = 1, r = 0: r ∨ q = 1, p ∨ (r ∨ q) = 1, p → q = 1, ¬r = 1, (p → q) ∨ ¬r = 1, (p ∨ (r ∨ q)) → ((p → q) ∨ ¬r) = 1
p = 0, q = 0, r = 1: r ∨ q = 1, p ∨ (r ∨ q) = 1, p → q = 1, ¬r = 0, (p → q) ∨ ¬r = 1, (p ∨ (r ∨ q)) → ((p → q) ∨ ¬r) = 1
p = 0, q = 0, r = 0: r ∨ q = 0, p ∨ (r ∨ q) = 0, p → q = 1, ¬r = 1, (p → q) ∨ ¬r = 1, (p ∨ (r ∨ q)) → ((p → q) ∨ ¬r) = 1

Count the rows with value 1

(p ∨ (r ∨ q)) → ((p → q) ∨ ¬r): 1, 1, 0, 1, 1, 1, 1, 1
N = 7

Answer

N = 7
4.Propositional logic & proofs

Answer: ¬r

Quantifiers and arguments: ¬∀x P(x) ≡ ∃x ¬P(x) and ¬∃x P(x) ≡ ∀x ¬P(x). Move a negation inwards, swap every quantifier it passes, and negate the inner formula with the propositional laws, such as ¬(a → b) ≡ a ∧ ¬b. An argument is valid when the conclusion is 1 in every row in which all premises are 1. A single row with true premises and a false conclusion is a counterexample and shows the argument is invalid.

Rows with all premises true

p = 0, q = 1, r = 0: p = 0, ¬r = 1, r = 0, r ∧ ¬p = 0

Conclusion

¬r (modus tollens with a conjunction)

Answer

¬r
5.Digital logic: gates, K-maps & flip-flops

Answer: N = 3

Karnaugh maps: the cells are ordered in Gray code (00, 01, 11, 10), so neighbouring cells, including the ones across the edges, differ in exactly one variable. Group the 1s in rectangles of 1, 2, 4, 8 or 16 cells, as large as possible and as few as possible, until every 1 is covered. Each group is one product term: the variables that stay constant inside it. A group of 2ᵏ cells loses k variables. Don't-cares (X) may be used to make groups bigger but never have to be covered. Grouping the 0s instead gives the minimal POS form.

ABC00011110010011031214150716

Group the 1s

Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.

B′C
BC′
AC′

Minimal SOP form

F = B′C + BC′ + AC′
N = 3

Answer

N = 3

Another grouping of the same size exists; it has the same number of terms and literals.

6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 12

Flip-flops and counters: a flip-flop changes only at the active clock edge. D copies its input (Q⁺ = D); T toggles when T = 1; JK holds for 00, resets for 01, sets for 10 and toggles for 11. An n-bit binary counter has 2ⁿ states and counts modulo 2ⁿ; a mod-N counter goes 0, 1, …, N − 1 and back to 0, and needs the smallest n with 2ⁿ ≥ N flip-flops. Each stage of a ripple counter halves the frequency, so the last output of a mod-N counter runs at fclk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.

State after every edge

1100 → 0110 → 0011 → 1001 → 1100

Decimal value

1100₂ = 12

Answer

Q = 12
7.Number theory & modular arithmetic

Answer: x = 70

Inverses and congruences: a has an inverse modulo n exactly when gcd(a, n) = 1. The extended Euclidean algorithm writes 1 = s·n + t·a, and then a⁻¹ ≡ t (mod n). A congruence a·x ≡ b (mod n) with gcd(a, n) = 1 has the solution x ≡ a⁻¹·b. Two congruences x ≡ r₁ (mod m₁) and x ≡ r₂ (mod m₂) with coprime moduli have one solution modulo m₁·m₂ (Chinese remainder theorem). Euler’s φ(n) counts the numbers from 1 to n that are coprime to n: φ(n) = n·∏(1 − 1/p).

Inverse of 65

r = 97, t = 0
r = 65, q = 1, t = 1
r = 32, q = 2, t = −1
r = 1, q = 32, t = 3
65−1 ≡ 3 (mod 97)

Solve

x ≡ 3·88 = 264 ≡ 70 (mod 97)
check: 65·70 = 4550 ≡ 88 (mod 97)

Answer

x = 70
8.Number theory & modular arithmetic

Answer: c = 107

Modular powers and RSA: compute aᵏ mod n by repeated squaring: write k in binary, square, reduce after every step and multiply the powers whose bit is 1. Fermat: aᵖ⁻¹ ≡ 1 (mod p) for a prime p not dividing a, so the exponent can be reduced modulo p − 1. RSA: choose primes p and q, n = p·q and φ(n) = (p − 1)(q − 1); the public exponent e is coprime to φ(n), and the private exponent is d = e⁻¹ mod φ(n). Encrypt with c = mᵉ mod n and decrypt with m = cᵈ mod n.

Repeated squaring

17 = 16 + 1
391 ≡ 39 (mod 187)
392 ≡ 25 (mod 187)
394 ≡ 64 (mod 187)
398 ≡ 169 (mod 187)
3916 ≡ 137 (mod 187)
3917 ≡ 39·137 ≡ 107 (mod 187)

Ciphertext

c = 107

Answer

c = 107
9.Algorithms & complexity

Answer: A · O(mn)

Why this is the answer

The table has (m + 1)(n + 1) cells and each is computed in O(1) from its neighbours.

Answer

A · O(mn)
10.Algorithms & complexity

Answer: C · In NP and every NP problem reduces to it in polynomial time

Why this is the answer

NP-complete problems are the hardest in NP; a polynomial algorithm for one would give one for all of NP.

Answer

C · In NP and every NP problem reduces to it in polynomial time
11.Data structures

Answer: D · Array elements are contiguous, so they make good use of the CPU cache

Why this is the answer

Contiguous memory gives good locality of reference, while linked nodes may be scattered and cause many cache misses.

Answer

D · Array elements are contiguous, so they make good use of the CPU cache
12.Data structures

Answer: C · To avoid repeated resizing when pushes and pops alternate near the boundary

Why this is the answer

Shrinking at one half could trigger a resize on every operation in such a sequence; the gap between 1/4 and full keeps the amortised cost O(1).

Answer

C · To avoid repeated resizing when pushes and pops alternate near the boundary
13.Graph theory

Answer: L = 20

Euler, Hamilton and trees: a connected graph has an Euler circuit (every edge once, back to the start) when every vertex has even degree, and an Euler path that is not a circuit when exactly two vertices have odd degree; the path then runs from one odd vertex to the other. A Hamilton circuit visits every vertex once; Kn has (n − 1)!/2 of them. A tree on n vertices has n − 1 edges, and a connected graph with m edges keeps a spanning tree after removing m − n + 1 edges.

Handshake lemma for a tree

L + 34 = 2(L + 8) − 2
L = 20

Answer

L = 20

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

14.Graph theory

Answer: N = 2

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 0 0 0 0
B: 1 0 1 1 0 0
C: 0 1 0 0 1 0
D: 0 1 0 0 1 1
E: 0 0 1 1 0 1
F: 0 0 0 1 1 0

One more step

(M³)EE = 0·0 + 2·0 + 0·1 + 1·1 + 3·0 + 1·1 = 2

Answer

N = 2
15.Databases & SQL

Answer: C · Choosing which columns appear in the result

Why this is the answer

Projection keeps chosen attributes (the SELECT list), while selection σ keeps rows that satisfy a condition (the WHERE clause).

Answer

C · Choosing which columns appear in the result
16.Databases & SQL

Answer: A · A recursive common table expression (WITH RECURSIVE)

Why this is the answer

A recursive CTE starts with an anchor query and repeatedly joins the previous result to the table, so it handles hierarchies of any depth.

Answer

A · A recursive common table expression (WITH RECURSIVE)
17.Computer networks

Answer: B · 254

Why this is the answer

A /24 has 8 host bits, so 2⁸ = 256 addresses. The network and broadcast addresses are reserved, leaving 254 for hosts.

Answer

B · 254
18.Computer networks

Answer: C · A lost TCP segment delays all multiplexed streams until it is retransmitted

Why this is the answer

TCP delivers bytes in order, so one loss stalls every stream on the connection. QUIC (HTTP/3) handles streams independently over UDP.

Answer

C · A lost TCP segment delays all multiplexed streams until it is retransmitted
19.IPv4 subnetting

Answer: x = 255

IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).

Block size

27 = 128

Subnet number 2

network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255

Answer

x = 255
20.IPv4 subnetting

Answer: x = 64

VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.

Block size per LAN

A: 34 hosts → 26 = 64 (/26)
D: 19 hosts → 25 = 32 (/27)
E: 12 hosts → 24 = 16 (/28)
B: 9 hosts → 24 = 16 (/28)
C: 6 hosts → 23 = 8 (/29)

Allocation, largest first

A: .0 – .63
D: .64 – .95
E: .96 – .111
B: .112 – .127
C: .128 – .135

Answer

x = 64

LAN D: 192.168.73.64/27.

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